Optimal. Leaf size=129 \[ \frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+3)}-\frac {(a-2 b (p+1)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )}{b f (2 p+3)} \]
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Rubi [A] time = 0.10, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4146, 388, 246, 245} \[ \frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+3)}-\frac {(a-2 b (p+1)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )}{b f (2 p+3)} \]
Antiderivative was successfully verified.
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Rule 245
Rule 246
Rule 388
Rule 4146
Rubi steps
\begin {align*} \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac {\operatorname {Subst}\left (\int \left (1+x^2\right ) \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac {(a-2 b (1+p)) \operatorname {Subst}\left (\int \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{b f (3+2 p)}\\ &=\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac {\left ((a-2 b (1+p)) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname {Subst}\left (\int \left (1+\frac {b x^2}{a+b}\right )^p \, dx,x,\tan (e+f x)\right )}{b f (3+2 p)}\\ &=\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac {(a-2 b (1+p)) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{b f (3+2 p)}\\ \end {align*}
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Mathematica [A] time = 2.26, size = 126, normalized size = 0.98 \[ \frac {\tan (e+f x) \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \left (a+b \sec ^2(e+f x)\right )^p \left ((2 b (p+1)-a) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a+b}\right )+\left (a+b \tan ^2(e+f x)+b\right ) \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^p\right )}{b f (2 p+3)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{4}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.74, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{4}\left (f x +e \right )\right ) \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\cos \left (e+f\,x\right )}^4} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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